1986 Distance Queries //LCA

Distance Queries

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 3708		Accepted: 1315
Case Time Limit: 1000MS

Description

Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in "Navigation Nightmare",followed by a line containing a single integer K, followed by K "distance queries". Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ's distance queries as quickly as possible!

Input

* Lines 1..1+M: Same format as "Navigation Nightmare"

* Line 2+M: A single integer, K. 1 <= K <= 10,000

* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.

Output

* Lines 1..K: For each distance query, output on a single line an integer giving the appropriate distance.

Sample Input

7 6
1 6 13 E
6 3 9 E
3 5 7 S
4 1 3 N
2 4 20 W
4 7 2 S
3
1 6
1 4
2 6

Sample Output

13
3
36

Hint

Farms 2 and 6 are 20+3+13=36 apart.

Source

USACO 2004 February

 

 

#include<stdio.h>//无向图，不知道根，离线查找两点间的距离，邻接表LCA
#include<string.h>
const int MAX=40005;
struct node
{
    int dis,first;
}point[MAX];
struct road
{
    int next,to,dis;
}rd[MAX*2],que[MAX*2];

int p[MAX],link[MAX],res[MAX];
bool vis[MAX];

void init(int n)//并查集
{
    for(int i=0;i<=n;i++)
    {
        p[i]=i;
        link[i]=-1;
        point[i].first=-1;
        res[i]=0;
        vis[i]=false;
    }
}
int find(int x)
{
    if(x!=p[x]) p[x]=find(p[x]);
    return p[x];
}
int Union(int a,int b)
{
    a=find(a);
    b=find(b);
    p[b]=a;
    return a;
}

void insertR(int from,int to,int len,int& n)
{
    rd[n].to=to;
    rd[n].dis=len;
    rd[n].next=point[from].first;
    point[from].first=n++;
}
void insertQ(int from,int to,int index,int& n)
{
    que[n].to=to;
 que[n].dis=index;
 que[n].next=link[from];
 link[from]=n++;
}

void lca(int pos,int dis)
{
    int p;
    point[pos].dis=dis;
    vis[pos]=true;
    for(int i=point[pos].first;i!=-1;i=rd[i].next)
    {
        if(!vis[rd[i].to])
        {
            lca(rd[i].to,dis+rd[i].dis);
            Union(pos,rd[i].to);
        }
    }
    for(int i=link[pos];i!=-1;i=que[i].next)
    {
        if(vis[que[i].to])
        {
            p=find(que[i].to);
            res[que[i].dis]=dis+point[que[i].to].dis-2*point[p].dis;//que[i].dis表示的是询问的序号
        }
    }
}

int main()
{
    int n,m,from,to,len,rn,qn,k;
    char c;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        rn=0,qn=0;
        init(n);//如果是多组数据，那么初始化要注意
        for(int i=0;i<m;i++)
        {
           scanf("%d %d %d %c",&from,&to,&len,&c);
           insertR(from,to,len,rn);//无向图
           insertR(to,from,len,rn);
        }
        scanf("%d",&k);
        for(int i=0;i<k;i++)
        {
            scanf("%d %d",&from,&to);
            insertQ(from,to,i,qn);
            insertQ(to,from,i,qn);
        }
        vis[1]=true;//假设一为根,如果是多组数组，注意数组的初始化
        lca(1,0);
        for(int i=0;i<k;i++) printf("%d/n",res[i]);
    }
    return 0;
}